Sederhanakanlah : a. ^2 log 32 - ^2 log 4 + ^2 log 1/8 b. ^7 log 4 + 2 ^7 log 3

Ganjar64 a.²log 32 - ²log 4 + ² log 1/8 = ²log (32:4x1/8)
   = ²log 1
   = 0
b.
⁷log 4 + 2 ⁷log 3 - 2 ⁷log 6 = ⁷log 4 + ⁷log 3² - ⁷log 6²
      = ⁷log 4 + ⁷log 9 - ⁷log 36
      = ⁷log (4x9:36)
      = ⁷log 1
      = 0

0 votes Thanks 1

IcukSugiarto $\bold{\underline{Solution\ a:}}\\~^{2}log\ 32 -~^{2}log\ 4+~^{2}log\ \frac{1}{8}\\\boxed{=~^{2}log\ \frac{32}{4} +~^{2}log\ \frac{1}{8} }\\\boxed{=~^{2}log\ 8 +~^{2}log\ \frac{1}{8} }\\\boxed{=~^{2}log\ (8\times \frac{1}{8}) }\\\boxed{=~^{2}log\ 1}\\\boxed{=~^{2}log\ 2^{0}}\\\boxed{=0 \times ~^{2}log\ 2}\\\boxed{=0 \times 1}\\\boxed{=0}$

$\bold{\underline{Solution\ b:}}\\~^{7}\log\ 4+2\ ~^{7}\log\ 3-2~^{7}\log\ 6\\\boxed{=~^{7}\log\ 4+ ~^{7}\log\ 3^{2}-~^{7}\log\ 6^{2}}\\\boxed{=~^{7}\log\ 4+ ~^{7}\log\ 9-~^{7}\log\ 36}\\\boxed{=~^{7}\log\ \frac{4\times 9}{36} }\\\boxed{=~^{7}\log\ \frac{36}{36} }\\\boxed{=~^{7}\log\ 1}\\\boxed{=~^{7}\log\ 7^{0}}\\\boxed{=0\cdot ~^{7}\log\ 7}\\\boxed{=0}$

$===================================\\\bold{Note:}\\\boxed{~^{a}\log\ (b\times c)=~^{a}\log\ b+~^{a}\log\ c}\\\boxed{~^{a}\log\ b^{m}=m~^{a}\log\ b}\\\boxed{~^{a}\log\ \frac{b}{c}=~^{a}\log\ b-~^{a}\log\ c }$

2 votes Thanks 1

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